this is an interesting remark. but if the internal value that creates
the microseconds value is also only a double, then there will not be
a rollover but a stop, because it cannot add 1 any more.
hypothetical, maybe it is much more precise internally.
matthias
On 31 août 06, at 18:30, Charles Yeomans wrote:
A correction -- it isn't rollover that will happen, but you'll
lose precision.
Charles Yeomans
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